Problem 5

We have seen that we can fit an SVM with a non-linear kernel in order to perform classification using a non-linear decision boundary. We will now see that we can also obtain a non-linear decision boundary by performing logistic regression using non-linear transformations of the features.

(a) Generate a data set with n = 500 and p = 2, such that the observations belong to two classes with a quadratic decision boundary between them.

set.seed(1)
x1 <- runif(500) - 0.5
x2 <- runif(500) - 0.5
y <- 1 * (x1^2 - x2^2 > 0)

(b) Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the yaxis.

plot(x1, x2, xlab = "X1", ylab = "X2", col = (4 - y), pch = (3 - y))

(c) Fit a logistic regression model to the data, using X1 and X2 as predictors.

logit.fit <- glm(y ~ x1 + x2, family = "binomial")
summary(logit.fit)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial")
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.179  -1.139  -1.112   1.206   1.257  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.087260   0.089579  -0.974    0.330
## x1           0.196199   0.316864   0.619    0.536
## x2          -0.002854   0.305712  -0.009    0.993
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 692.18  on 499  degrees of freedom
## Residual deviance: 691.79  on 497  degrees of freedom
## AIC: 697.79
## 
## Number of Fisher Scoring iterations: 3

The variables are not statistically significant.

(d) Apply this model to the training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be linear.

data <- data.frame(x1 = x1, x2 = x2, y = y)
probs <- predict(logit.fit, data, type = "response")
preds <- rep(0, 500)
preds[probs > 0.47] <- 1
plot(data[preds == 1, ]$x1, data[preds == 1, ]$x2, col = (4 - 1), pch = (3 - 1), xlab = "X1", ylab = "X2")
points(data[preds == 0, ]$x1, data[preds == 0, ]$x2, col = (4 - 0), pch = (3 - 0))

(e) Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X21 , X1×X2, log(X2), and so forth).

logitnl.fit <- glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), family = "binomial")
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
summary(logitnl.fit)
## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), family = "binomial")
## 
## Deviance Residuals: 
##        Min          1Q      Median          3Q         Max  
## -8.240e-04  -2.000e-08  -2.000e-08   2.000e-08   1.163e-03  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept)    -102.2     4302.0  -0.024    0.981
## poly(x1, 2)1   2715.3   141109.5   0.019    0.985
## poly(x1, 2)2  27218.5   842987.2   0.032    0.974
## poly(x2, 2)1   -279.7    97160.4  -0.003    0.998
## poly(x2, 2)2 -28693.0   875451.3  -0.033    0.974
## I(x1 * x2)     -206.4    41802.8  -0.005    0.996
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9218e+02  on 499  degrees of freedom
## Residual deviance: 3.5810e-06  on 494  degrees of freedom
## AIC: 12
## 
## Number of Fisher Scoring iterations: 25

The variables are not statistically significant.

(f) Apply this model to the training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be obviously non-linear. If it is not, then repeat (a)-(e) until you come up with an example in which the predicted class labels are obviously non-linear.

probs <- predict(logitnl.fit, data, type = "response")
preds <- rep(0, 500)
preds[probs > 0.47] <- 1
plot(data[preds == 1, ]$x1, data[preds == 1, ]$x2, col = (4 - 1), pch = (3 - 1), xlab = "X1", ylab = "X2")
points(data[preds == 0, ]$x1, data[preds == 0, ]$x2, col = (4 - 0), pch = (3 - 0))

(g) Fit a support vector classifier to the data with X1 and X2 as predictors. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

library(e1071)
data$y <- as.factor(data$y)
svm.fit <- svm(y ~ x1 + x2, data, kernel = "linear", cost = 0.01)
preds <- predict(svm.fit, data)
plot(data[preds == 0, ]$x1, data[preds == 0, ]$x2, col = (4 - 0), pch = (3 - 0), xlab = "X1", ylab = "X2")
points(data[preds == 1, ]$x1, data[preds == 1, ]$x2, col = (4 - 1), pch = (3 - 1))

(h) Fit a SVM using a non-linear kernel to the data. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

data$y <- as.factor(data$y)
svmnl.fit <- svm(y ~ x1 + x2, data, kernel = "radial", gamma = 1)
preds <- predict(svmnl.fit, data)
plot(data[preds == 0, ]$x1, data[preds == 0, ]$x2, col = (4 - 0), pch = (3 - 0), xlab = "X1", ylab = "X2")
points(data[preds == 1, ]$x1, data[preds == 1, ]$x2, col = (4 - 1), pch = (3 - 1))

(i) Comment on your results.

We find how important it is to use SVMS for finding non linear models.

Problem 7

In this problem, you will use support vector approaches in order to predict whether a given car gets high or low gas mileage based on the Auto data set.

(a) Create a binary variable that takes on a 1 for cars with gas mileage above the median, and a 0 for cars with gas mileage below the median.

library(ISLR)
## Warning: package 'ISLR' was built under R version 3.6.2
gas.med = median(Auto$mpg)
new.var = ifelse(Auto$mpg > gas.med, 1, 0)
Auto$mpglevel = as.factor(new.var)

(b) Fit a support vector classifier to the data with various values of cost, in order to predict whether a car gets high or low gas mileage. Report the cross-validation errors associated with different values of this parameter. Comment on your results.

set.seed(3255)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 
    0.1, 1, 5, 10, 100)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.01269231 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-02 0.07397436 0.06863413
## 2 1e-01 0.05102564 0.06923024
## 3 1e+00 0.01269231 0.02154160
## 4 5e+00 0.01519231 0.01760469
## 5 1e+01 0.02025641 0.02303772
## 6 1e+02 0.03294872 0.02898463

Cost of 1 performs best.

(c) Now repeat (b), this time using SVMs with radial and polynomial basis kernels, with different values of gamma and degree and cost. Comment on your results.

set.seed(1)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.1, 
    1, 5, 10), degree = c(2, 3, 4)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##    10      2
## 
## - best performance: 0.5130128 
## 
## - Detailed performance results:
##    cost degree     error dispersion
## 1   0.1      2 0.5511538 0.04366593
## 2   1.0      2 0.5511538 0.04366593
## 3   5.0      2 0.5511538 0.04366593
## 4  10.0      2 0.5130128 0.08963366
## 5   0.1      3 0.5511538 0.04366593
## 6   1.0      3 0.5511538 0.04366593
## 7   5.0      3 0.5511538 0.04366593
## 8  10.0      3 0.5511538 0.04366593
## 9   0.1      4 0.5511538 0.04366593
## 10  1.0      4 0.5511538 0.04366593
## 11  5.0      4 0.5511538 0.04366593
## 12 10.0      4 0.5511538 0.04366593

With polynomial kernel, the lowest cross-validation error is obtained for a degree of 2 and a cost of 10.

set.seed(1)
tune.out = tune(svm, mpglevel ~ ., data = Auto, kernel = "radial", ranges = list(cost = c(0.1, 
    1, 5, 10), gamma = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##    10  0.01
## 
## - best performance: 0.02557692 
## 
## - Detailed performance results:
##    cost gamma      error dispersion
## 1   0.1 1e-02 0.08929487 0.04382379
## 2   1.0 1e-02 0.07403846 0.03522110
## 3   5.0 1e-02 0.04852564 0.03303346
## 4  10.0 1e-02 0.02557692 0.02093679
## 5   0.1 1e-01 0.07903846 0.03874545
## 6   1.0 1e-01 0.05371795 0.03525162
## 7   5.0 1e-01 0.02820513 0.03299190
## 8  10.0 1e-01 0.03076923 0.03375798
## 9   0.1 1e+00 0.55115385 0.04366593
## 10  1.0 1e+00 0.06384615 0.04375618
## 11  5.0 1e+00 0.05884615 0.04020934
## 12 10.0 1e+00 0.05884615 0.04020934
## 13  0.1 5e+00 0.55115385 0.04366593
## 14  1.0 5e+00 0.49493590 0.04724924
## 15  5.0 5e+00 0.48217949 0.05470903
## 16 10.0 5e+00 0.48217949 0.05470903
## 17  0.1 1e+01 0.55115385 0.04366593
## 18  1.0 1e+01 0.51794872 0.05063697
## 19  5.0 1e+01 0.51794872 0.04917316
## 20 10.0 1e+01 0.51794872 0.04917316
## 21  0.1 1e+02 0.55115385 0.04366593
## 22  1.0 1e+02 0.55115385 0.04366593
## 23  5.0 1e+02 0.55115385 0.04366593
## 24 10.0 1e+02 0.55115385 0.04366593

With radial kernel, the lowest cross-validation error is obtained for a gamma of 0.01 and a cost of 10.

(d) Make some plots to back up your assertions in (b) and (c). Hint: In the lab, we used the plot() function for svm objects only in cases with p = 2. When p > 2, you can use the plot() function to create plots displaying pairs of variables at a time. Essentially, instead of typing > plot(svmfit , dat) where svmfit contains your fitted model and dat is a data frame containing your data, you can type > plot(svmfit , dat , x1∼x4) in order to plot just the first and fourth variables. However, you must replace x1 and x4 with the correct variable names. To find out more, type ?plot.svm.

svm.linear = svm(mpglevel ~ ., data = Auto, kernel = "linear", cost = 1)
svm.poly = svm(mpglevel ~ ., data = Auto, kernel = "polynomial", cost = 10, 
    degree = 2)
svm.radial = svm(mpglevel ~ ., data = Auto, kernel = "radial", cost = 10, gamma = 0.01)
plotpairs = function(fit) {
    for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpglevel", "name"))]) {
        plot(fit, Auto, as.formula(paste("mpg~", name, sep = "")))
    }
}
plotpairs(svm.linear)

Problem 8

This problem involves the OJ data set which is part of the ISLR package.

(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

set.seed(1)
train <- sample(nrow(OJ), 800)
OJ.train <- OJ[train, ]
OJ.test <- OJ[-train, ]

(b) Fit a support vector classifier to the training data using cost=0.01, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.

svm.linear <- svm(Purchase ~ ., data = OJ.train, kernel = "linear", cost = 0.01)
summary(svm.linear)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "linear", 
##     cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  435
## 
##  ( 219 216 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

With the summary statistics we find the support vector classifier creates 435 support vectors of the 800 training points. Out of the 435 support vectors, 219 belong to level CH and the remaining 216 belong to level MM.

(c) What are the training and test error rates?

train.pred <- predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 420  65
##   MM  75 240
(75 + 65) / (420 + 240 + 75 + 65)
## [1] 0.175
test.pred <- predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 153  15
##   MM  33  69
(33 + 15) / (153 + 69 + 33 + 15)
## [1] 0.1777778

The training error rate is 17.5% and test error rate is about 17.77%.

(d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.

set.seed(2)
tune.out <- tune(svm, Purchase ~ ., data = OJ.train, kernel = "linear", ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  1.778279
## 
## - best performance: 0.1675 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.17625 0.04059026
## 2   0.01778279 0.17625 0.04348132
## 3   0.03162278 0.17125 0.04604120
## 4   0.05623413 0.17000 0.04005205
## 5   0.10000000 0.17125 0.04168749
## 6   0.17782794 0.17000 0.04090979
## 7   0.31622777 0.17125 0.04411554
## 8   0.56234133 0.17125 0.04084609
## 9   1.00000000 0.17000 0.04090979
## 10  1.77827941 0.16750 0.03782269
## 11  3.16227766 0.16750 0.03782269
## 12  5.62341325 0.16750 0.03545341
## 13 10.00000000 0.17000 0.03736085

Optimal cost is 1.778279.

(e) Compute the training and test error rates using this new value for cost.

svm.linear <- svm(Purchase ~ ., kernel = "linear", data = OJ.train, cost = tune.out$best.parameter$cost)
train.pred <- predict(svm.linear, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 423  62
##   MM  69 246
(69 + 62) / (423 + 246 + 69 + 62)
## [1] 0.16375
test.pred <- predict(svm.linear, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 156  12
##   MM  29  73
(29 + 12) / (156 + 73 + 29 + 12)
## [1] 0.1518519

With the best cost, the training error rate is now 16.37% and the test error rate is 15.18%.

(f) Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.

svm.radial <- svm(Purchase ~ ., kernel = "radial", data = OJ.train)
summary(svm.radial)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "radial")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  373
## 
##  ( 188 185 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred <- predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 441  44
##   MM  77 238
(77 + 44) / (441 + 238 + 77 + 44)
## [1] 0.15125
test.pred <- predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 151  17
##   MM  33  69
(33 + 17) / (151 + 69 + 33 + 17)
## [1] 0.1851852

Radial kernel with default gamma creates 373 support vectors, out of which, 188 belong to level CH and remaining 185 belong to level MM. The classifier has a training error of 15.12% and a test error of 18.51% which is a slight improvement over linear kernel.

set.seed(2)
tune.out <- tune(svm, Purchase ~ ., data = OJ.train, kernel = "radial", ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.1725 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39375 0.03240906
## 2   0.01778279 0.39375 0.03240906
## 3   0.03162278 0.34750 0.05552777
## 4   0.05623413 0.19250 0.03016160
## 5   0.10000000 0.19500 0.03782269
## 6   0.17782794 0.18000 0.04048319
## 7   0.31622777 0.17250 0.03809710
## 8   0.56234133 0.17500 0.04124790
## 9   1.00000000 0.17250 0.03162278
## 10  1.77827941 0.17750 0.03717451
## 11  3.16227766 0.18375 0.03438447
## 12  5.62341325 0.18500 0.03717451
## 13 10.00000000 0.18750 0.03173239
svm.radial <- svm(Purchase ~ ., kernel = "radial", data = OJ.train, cost = tune.out$best.parameter$cost)
summary(svm.radial)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "radial", 
##     cost = tune.out$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  373
## 
##  ( 188 185 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred <- predict(svm.radial, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 441  44
##   MM  77 238
(77 + 44) / (441 + 238 + 77 + 44)
## [1] 0.15125
test.pred <- predict(svm.radial, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 151  17
##   MM  33  69
(33 + 17) / (151 + 69 + 33 + 17)
## [1] 0.1851852

We find that tuning does not reduce the train and test error rates.

(g) Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.

svm.poly <- svm(Purchase ~ ., kernel = "polynomial", data = OJ.train, degree = 2)
summary(svm.poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "polynomial", 
##     degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  447
## 
##  ( 225 222 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred <- predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 449  36
##   MM 110 205
(110 + 36) / (449 + 205 + 110 + 36)
## [1] 0.1825
test.pred <- predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 153  15
##   MM  45  57
(45 + 15) / (153 + 57 + 45 + 15)
## [1] 0.2222222

Polynomial kernel with default gamma creates 447 support vectors, out of which, 225 belong to level CH and remaining 222 belong to level MM. The classifier has a training error of 18.25% and a test error of 22.22% which is no improvement over linear kernel.

set.seed(2)
tune.out <- tune(svm, Purchase ~ ., data = OJ.train, kernel = "polynomial", degree = 2, ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(tune.out)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  3.162278
## 
## - best performance: 0.18 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39000 0.03670453
## 2   0.01778279 0.37000 0.03395258
## 3   0.03162278 0.36375 0.03197764
## 4   0.05623413 0.34500 0.03291403
## 5   0.10000000 0.32125 0.03866254
## 6   0.17782794 0.24750 0.03322900
## 7   0.31622777 0.20250 0.04073969
## 8   0.56234133 0.20250 0.03670453
## 9   1.00000000 0.19625 0.03910900
## 10  1.77827941 0.19125 0.03586723
## 11  3.16227766 0.18000 0.04005205
## 12  5.62341325 0.18000 0.04133199
## 13 10.00000000 0.18125 0.03830162
svm.poly <- svm(Purchase ~ ., kernel = "polynomial", degree = 2, data = OJ.train, cost = tune.out$best.parameter$cost)
summary(svm.poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "polynomial", 
##     degree = 2, cost = tune.out$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  3.162278 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  385
## 
##  ( 197 188 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
train.pred <- predict(svm.poly, OJ.train)
table(OJ.train$Purchase, train.pred)
##     train.pred
##       CH  MM
##   CH 451  34
##   MM  90 225
(90 + 34) / (451 + 225 + 90 + 34)
## [1] 0.155
test.pred <- predict(svm.poly, OJ.test)
table(OJ.test$Purchase, test.pred)
##     test.pred
##       CH  MM
##   CH 154  14
##   MM  41  61
(41 + 14) / (154 + 61 + 41 + 14)
## [1] 0.2037037

Tuning reduced both the train and test errors.

(h) Overall, which approach seems to give the best results on this data? Overall, the radial basis kernel seems to give the best results as it produced minimum misclassification errors on both test and train data.